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Question Number 33496 by malwaan last updated on 17/Apr/18

prove that  e^(iπ) +1=0

$$\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{e}^{\mathrm{i}\pi} +\mathrm{1}=\mathrm{0} \\ $$

Answered by MJS last updated on 17/Apr/18

just use the definition of z∈C:  z=re^(ϕi) =rcos(ϕ)+rsin(ϕ)i ⇒  ⇒ e^(πi) =1cos(π)+1sin(π)i=  =−1+0i=−1  −1+1=0

$$\mathrm{just}\:\mathrm{use}\:\mathrm{the}\:\mathrm{definition}\:\mathrm{of}\:{z}\in\mathbb{C}: \\ $$$${z}={r}\mathrm{e}^{\varphi\mathrm{i}} ={r}\mathrm{cos}\left(\varphi\right)+{r}\mathrm{sin}\left(\varphi\right)\mathrm{i}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{e}^{\pi\mathrm{i}} =\mathrm{1cos}\left(\pi\right)+\mathrm{1sin}\left(\pi\right)\mathrm{i}= \\ $$$$=−\mathrm{1}+\mathrm{0i}=−\mathrm{1} \\ $$$$−\mathrm{1}+\mathrm{1}=\mathrm{0} \\ $$

Commented by malwaan last updated on 18/Apr/18

thank you so much

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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